Organic Chemistry 1 · Stereochemistry of alkene reactions
Butanone and acetaldehyde from (Z)-3-methylpent-2-ene
Conditions: O3, CH2Cl2, −78 °C; then (CH3)2S
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The mechanism, step by step
Next plays each step's curved arrows and explains them. The written steps are below.
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Add ozone across the C=C
Ozone adds across the trisubstituted C=C of (Z)-3-methylpent-2-ene in one concerted [3+2] cycloaddition. The Z arrangement of the substituents is kept in the molozonide for the moment.
The curved arrows
- A lone pair on the negatively charged end oxygen of ozone bonds to the alkene carbon bearing the methyl and ethyl groups.
- The C=C π pair bonds the alkene CH carbon to the double-bonded end oxygen of ozone.
- The O=O π pair of ozone moves onto the central oxygen, which loses its positive charge.
Split the molozonide
The molozonide fragments in a reverse [3+2], breaking the C–C bond. The CH end leaves as acetaldehyde, and the carbon carrying the methyl and ethyl groups keeps the carbonyl oxide; at this point the original geometry is gone.
The curved arrows
- The old C–C σ pair becomes a C=O π bond at the alkene CH carbon, which is bonded to an end oxygen.
- The weak O–O bond next to it breaks and its pair moves onto the central oxygen, which becomes negative.
- A lone pair on the other end oxygen forms a C=O⁺ π bond to the alkene carbon bearing the methyl and ethyl groups, giving the carbonyl oxide.
Recombine to the ozonide
The carbonyl oxide is itself a 1,3-dipole, so it adds back across the C=O group in the opposite orientation. This second [3+2] gives the ozonide, a 1,2,4-trioxolane with just one O–O bond.
The curved arrows
- The negative oxygen of the carbonyl oxide bonds to the carbonyl carbon that came from the alkene CH carbon.
- The C=O π pair bonds that carbonyl oxygen to the carbonyl-oxide carbon from the alkene carbon bearing the methyl and ethyl groups.
- The C=O⁺ π pair returns to its oxygen, which becomes neutral again, closing the five-membered ozonide.
Dimethyl sulfide opens the peroxide
Dimethyl sulfide attacks a peroxide oxygen of the ozonide and breaks the O–O bond, leaving an alkoxide on the carbon that carries the methyl and ethyl groups.
The curved arrows
- A lone pair on sulfur attacks one oxygen of the ozonide's peroxide bond.
- The weak O–O pair moves onto the other peroxide oxygen, which becomes an alkoxide.
Release the first carbonyl
That alkoxide re-forms C=O and expels the ring oxygen, releasing butanone.
The curved arrows
- The alkoxide lone pair pushes back down to form a C=O π bond.
- The C–O bond to the ring oxygen breaks, and that oxygen leaves as a new alkoxide.
Release the second carbonyl and DMSO
The second alkoxide forms the aldehyde C=O and pushes out the sulfur-bound oxygen as DMSO, releasing acetaldehyde.
The curved arrows
- The new alkoxide lone pair forms the second C=O π bond.
- The C–O bond to the sulfur-bound oxygen breaks, releasing dimethyl sulfoxide (DMSO).
E or Z, same products?
Ozonolysis breaks the C=C, so the products cannot remember which side each group was on.
Acetaldehyde and butanone
Cleavage gives butanone from the carbon with the methyl and ethyl groups and acetaldehyde from the CH end. Because both alkene carbons become flat carbonyl carbons, the Z and E isomers give exactly the same pair of products.
What you'll practice
Draw the ozonolysis of (Z)-3-methylpent-2-ene with a dimethyl sulfide workup and explain why its E isomer would give the same butanone and acetaldehyde.
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